Calculate the pOH of a solution of methylamine and lye. Initially, [CH3NH2] = 1M and [NaOH] = 0.02M. CH3NH2(aq) + H2O(l) ↔ CH3NH3+(aq) + OH-(aq)      Kb = 4.38x10-4 NaOH(aq) → Na+(aq) + OH-(aq)      Strong base   a. 1.70 b. 1.39 c. 1.68 d. 4.12 e. 2.33

Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter14: Acids And Bases
Section14.8: Acid-base Reactions Of Salts
Problem 14.21CE
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Calculate the pOH of a solution of methylamine and lye. Initially, [CH3NH2] = 1M and [NaOH] = 0.02M.

CH3NH2(aq) + H2O(l) ↔ CH3NH3+(aq) + OH-(aq)      Kb = 4.38x10-4

NaOH(aq) → Na+(aq) + OH-(aq)      Strong base

 

a.

1.70

b.

1.39

c.

1.68

d.

4.12

e.

2.33

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